2 stroke thread (with occasional F1 relevance!)

All that has to do with the power train, gearbox, clutch, fuels and lubricants, etc. Generally the mechanical side of Formula One.
manolis
73
Joined: Tue Mar 18, 2014 9:00 am

Re: 2 stroke thread (with occasional F1 relevance!)

Post by manolis » Sat Aug 05, 2017 5:21 am

Hello Pinger.

You write:
“..... and now ponder how to emulate a scooter CVT transmitting circa 5lbs.ft of torque with a 200lb cyclist generating 100lbs.ft torque without the belt slipping. Manipulate the force required for that from the handlebar? It'd be worse than tensioning a truck's fan belt - while simultaneously riding a bike.”



I don’t know what you mean by “tensioning a truck's fan belt”.

So, let’s make a rough calculation to find out how hard it is to shift to another gear ratio with the lever of the PatBox in a bicycle.


Suppose the coefficient of friction between the elastic V-belt and the conical pulley halves equals to 1.

Suppose a 100Kp (220lb) bicycler weight.

Suppose a 150mm (0.5ft) pedal crank-arm.

Suppose a 75mm (0.25ft) maximum radius of the V-belt running on the rear pulley, and a 1:1 minimum transmission ratio.

150mm/75mm=2

So, the rear pulley, at the shortest gear ratio (i.e. at the ratio the bicycle starts moving), has to apply a force of 2*100=200Kp (440lb) on the V-belt, so it is required a clamping force of 100Kp (220lb); (because force is applied on both sides of the V-belt).

In order the auxiliary belt to displace the V-belt deeper in the rear pulley, it is required to apply to the V-belt a force of 100/4=25Kp (55lb): this is due to the wedge effect and to the angle of the conical pulley halves (say 28 degrees, tan(28/2)=0.25=1/4).

The auxiliary belt has two spans, one above and another below the V-belt. So, each span of the auxiliary belt has to be loaded by a force of 25/2=12.5Kp (28lb).

This is the force the movable roller-tensioner has to apply to the auxiliary belt horizontally.

Image

The handle (the top) of the lever is at a 3.5:1, or so, “leverage” relative to the upper roller-tensioner, which means, the hand of the bicycler has to apply on the lever a force of 12.5/3.5 = 3.5Kp (8lb), i.e. as much as a lightweight domestic cat weighs (from the Internet: according to APOP, a domestic cat should be about 8-10 lbs (3.6-4.5 kg)).


The previous rough calculations give an idea for the magnitude of the required force on the lever.
And it seems OK not only for the average bicycler, but also for the extra lightweight bicyclers.


The arrangement of the PatBox lever may seem strange and unconventional to the familiar with the derailleur / IGH / NuVinci bicycle gear-trains and control.

However for the new bicyclers, and for the bicyclers who dislike, or never got how to use efficiently, the conventional bicycle gear-train and control, the lever of the PatBox appears quite more physical and simple and understandable.


The following instructions for the use of the PatBox:

“When your legs are pedaling too fast push the lever forwards, and when your legs are pedaling too slow pull the lever backwards.”

is all a bicycler needs to know.

Thanks
Manolis Pattakos

J.A.W.
48
Joined: Mon Sep 01, 2014 4:10 am
Location: Altair IV.

Re: 2 stroke thread (with occasional F1 relevance!)

Post by J.A.W. » Wed Aug 09, 2017 10:59 am

Hi Manolis..

Back on topic with 2T tech, but still of interest, with a - 'twisted' - belt drive, rotary-valve set-up..

https://www.youtube.com/watch?v=7Ffs1x145zw
Dr Everett V. Scott sez.. " I happen to know a great deal about a lot of things."

manolis
73
Joined: Tue Mar 18, 2014 9:00 am

Re: 2 stroke thread (with occasional F1 relevance!)

Post by manolis » Wed Aug 09, 2017 5:50 pm

Hello J.A.W.

The diameter of the rotary valve seems quite big (200mm?) for the size of the engine.

And the diameter of the sprockets whereon the twisted tooth belt runs, seems too small (on the other hand, in case the toothed belt fails, the engine just stalls).

It seems he is trying to achieve both: bigger intake port area and faster opening - closing.
Maybe he is also trying to avoid the limitations from the reed valve.

The next reasonable step for the optimization of the intake in a wide rev range? A phaser.



By the way, the following "animation" shows a phaser (like those used for the timing of the camshafts of the 4-stroke engines) disposed between the crankshaft and the bevel gear that drives the rotary valve of the PatATE:

Image

Depending on the operational conditions, the controllable timing can give bigger share of the hybrid port time area to the exhaust, or to the transfer.

It can also optimize the intake (at lower revs the intake port should close earlier, at higher revs the intake port should close later).

Thanks
Manolis Pattakos

manolis
73
Joined: Tue Mar 18, 2014 9:00 am

Re: 2 stroke thread (with occasional F1 relevance!)

Post by manolis » Sun Aug 13, 2017 5:34 am

Hello all.


PatCVT

In the PatCVT a sprocket or gearwheel (actually an idler) is intermeshed with "teeth" on both spans of the V-belt.
The displacement of the center of the sprocket varies positively the transmission ratio:

Image

enabling various "modes" of operation:

Drive Mode:
The lever that holds the sprocket is released to pivot freely about the cross; the CVT runs automatically under the control of a variator in the one conical pulley and of a spring / torque cam in the other conical pulley.

Sport Mode:
Performance oriented.
A control spring pushes the lever towards the conical pulley with the spring / torque cam; the CVT continues to run "fully automatic", but it selects shorter transmission ratios.

Economy Mode:
For green, quiet, reliable, comfortable operation.
The control spring pushes the lever towards the variator's conical pulley; the CVT continues to run "fully automatic" (as in the "Drive Mode"), but it selects longer transmission ratios.

Manual Mode:
The rider / driver displaces "manually" the lever about the cross to select any transmission ratio from the available infinity, regardless of what the "variator / spring / torque cam" command.
By locking the lever at a number of discrete positions, the CVT replicates a manual gear box.


In all modes the sprocket acts as a "baffle roller" (drive belt tensioner), too.

The available space inside the CVT of many scooters (like, for instance, the Sport-City Aprilia 250cc, below) is more than what the PatCVT needs:

Image


How it works?

With the sprocket engaged with the toothed-V-belt as in the animation, the sprocket divides the V-belt in two parts (one at left of the sprocket and another at right of the sprocket), with each part maintaining its length constant.

The move away of the center of the sprocket from the one conical pulley causes the decrease of the effective diameter the V-belt is running on this conical pulley, and the increase of the effective diameter the V-belt is running on the other conical pulley.



For more: http://www.pattakon.com/pattakonPatCVT.htm


Thoughts?

Thanks
Manolis Pattakos

Tommy Cookers
364
Joined: Fri Feb 17, 2012 3:55 pm

Re: 2 stroke thread (with occasional F1 relevance!)

Post by Tommy Cookers » Sun Aug 13, 2017 10:33 am

at this moment I seem to need convincing that the argument below is incorrect (and how/why it is incorrect) .......

in the PatBox the controlling belt has rather small losses (that might be smaller again if it was eg a metal band)
in the PatCVT the controlling belt has loads passing through its teeth via the pulley teeth ? (the belt teeth you have said before are not intended for loads)
because the belt load path on the other side of the control point is 'length-compliant' (not rigid as in the PatBox)

so the controlling losses seem to be higher in the PatCVT

what can the PatCVT do that couldn't be done by controlling belt effective length/tension with an external tensioner ? (eg roller on back side of belt)
ie is its benefit entirely its compact latout ?

and the question (of whether losses are greater than in conventional systems of external CVT control) is still open ?

it might be argued that the Pat CVT is an enabler of the 2 stroke 'daily driver'

manolis
73
Joined: Tue Mar 18, 2014 9:00 am

Re: 2 stroke thread (with occasional F1 relevance!)

Post by manolis » Sun Aug 13, 2017 4:01 pm

Hello Tommy Cookers.

You write:
“in the PatCVT the controlling belt has loads passing through its teeth via the pulley teeth ? (the belt teeth you have said before are not intended for loads)
because the belt load path on the other side of the control point is 'length-compliant' (not rigid as in the PatBox)”


In the PatCVT there is no “controlling belt”.
There is only one belt, the V-belt, that has teeth at its inner side.

Image

The controller is the sprocket intermeshed with teeth on both spans of the V-belt.


Regarding the efficiency:

Take a scooter CVT, like, say, the SprotCity Aprilia 250cc of the photo, modified to PatCVT.

If you release the lever that holds the controlling sprocket, the bearing of the sprocket runs, more or less, unloaded; the teeth of the sprocket remain unloaded, too.
The sprocket acts as the original “baffle” roller.
The efficiency of the CVT is as without the PatCVT.

Now suppose that at some conditions you prefer a 15% shorter, or longer, transmission ratio than the ratio the CVT selects normally at these conditions.
The bearing of the sprocket and its teeth are far from being overloaded, because the sprocket has to apply to the V-belt only the difference of the force required for the specific transmission ratio, and because the efficiency of a toothed sprocket cooperating with a toothed belt is quite high (above 99%; worth to note: there is only one sprocket).

For instance, suppose a force of 20Kp (44lb) is applied on the center of the sprocket towards the one conical pulley; this force changes by 10Kp (22lb) the force acting on each span of the V-belt; and due to the wedge action of the conical pulley, this force (the 20Kp / 44lb) is multiplied by 4 to give the additional axial force resulting on the moving half of each conical pulley.
So, with 20Kp loading the roller bearing of the sprocket, and 10Kp force from the sprocket on each span of the V-belt, the axial force on the movable halve of each conical pulleys change by 80Kp (176 lb).

If something of the previous is confusing, please let me know to further explain.



You also write:
“what can the PatCVT do that couldn't be done by controlling belt effective length/tension with an external tensioner ? (eg roller on back side of belt)
ie is its benefit entirely its compact latout ?”

Any drawing?
I can’t get what you mean.

Thanks
Manolis Pattakos

Tommy Cookers
364
Joined: Fri Feb 17, 2012 3:55 pm

Re: 2 stroke thread (with occasional F1 relevance!)

Post by Tommy Cookers » Mon Aug 14, 2017 10:51 am

you say .....
'the sprocket has to apply to the V belt only the difference in the forces'

I think losses will be produced by belt/sprocket action associated with each of the forces
ie proportionate to the algebraic sum of the forces not the difference in forces

manolis
73
Joined: Tue Mar 18, 2014 9:00 am

Re: 2 stroke thread (with occasional F1 relevance!)

Post by manolis » Mon Aug 14, 2017 2:47 pm

Hello Tommy Cookers.

You write:
“I think losses will be produced by belt/sprocket action associated with each of the forces
ie proportionate to the algebraic sum of the forces not the difference in forces”


Let’s take the marginal case wherein the lever of the sprocket is released to move freely.

The force between the sprocket and each span of the V-belt is near zero.
The force on the bearing of the sprocket is small, too.
These make the losses “produced by the belt/sprocket action” actually zero.

Obviously the strong force that loads the V-belt (through this force the one conical pulley transmits power to the other conical pulley) does not affect the above, almost zero, friction losses.

Do I miss something?

Thanks
Manolis Pattakos

Tommy Cookers
364
Joined: Fri Feb 17, 2012 3:55 pm

Re: 2 stroke thread (with occasional F1 relevance!)

Post by Tommy Cookers » Wed Aug 16, 2017 10:23 am

since you ask, I find myself saying that you are missing a response to my point
and you are pre-emptively trivialising the loss aspect of the forces developed at the belt/control pulley contact region

however I wish your designs success in their widening of the scope of inexpensive properly controlled CVTs



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